Problem: Multiply the following complex numbers: $({3-4i}) \cdot ({-3i})$
Answer: Complex numbers are multiplied like any two binomials. First use the distributive property: $ ({3-4i}) \cdot ({-3i}) = $ $ ({3} \cdot {0}) + ({3} \cdot {-3}i) + ({-4}i \cdot {0}) + ({-4}i \cdot {-3}i) $ Then simplify the terms: $ (0) + (-9i) + (0i) + (12 \cdot i^2) $ Imaginary unit multiples can be grouped together. $ 0 + (-9 + 0)i + 12i^2 $ After we plug in $i^2 = -1$ , the result becomes $ 0 + (-9 + 0)i - 12 $ The result is simplified: $ (0 - 12) + (-9i) = -12-9i $